The average travel time to work for a person living and working in Kokomo, Indiana is 19 minutes. Suppose the standard deviation of travel time to work is 4.5 minutes and the distribution of travel time is approximately normally distributed.
What is the travel time to work that separates the bottom 2.5% of people with the median or 50%? Round to the nearest tenth of a minute.
you can play around with Z-table stuff here:
http://davidmlane.com/hyperstat/z_table.html
To find the travel time that separates the bottom 2.5% of people with the median or 50%, we need to determine the z-score corresponding to that percentile.
A z-score measures how many standard deviations an observation is from the mean in a standard normal distribution. We can calculate the z-score using the formula:
z = (x - μ) / σ
Where:
z = z-score
x = travel time to work
μ = mean travel time to work
σ = standard deviation of travel time to work
In this case, the mean travel time to work is 19 minutes and the standard deviation is 4.5 minutes. Let's plug in the values into the formula:
z = (x - 19) / 4.5
To find the z-score that corresponds to the bottom 2.5% (or the left-tail area), we need to find the z-score that leaves 2.5% in the right tail, which is 100% - 2.5% = 97.5%. We can look up this z-score in a standard normal distribution table or use a calculator.
Using a standard normal distribution table, we find that the z-score corresponding to 97.5% is approximately 1.96.
Now, we can rearrange the formula to solve for x:
x = (z * σ) + μ
x = (1.96 * 4.5) + 19
x ≈ 8.82 + 19
x ≈ 27.82
Therefore, the travel time to work that separates the bottom 2.5% of people with the median or 50% is approximately 27.8 minutes (rounded to the nearest tenth of a minute).